Question: $\lim_{x\to\infty}\dfrac{8-3x}{9x^2+11}=?$ Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{1}{3}$ (Choice B) B $0$ (Choice C) C $\dfrac{8}{9}$ (Choice D) D $-\infty$
Answer: $\lim_{x\to\infty} 8-3x=-\infty$ and $\lim_{x\to\infty} 9x^2+11=\infty$, so $\lim_{x\to\infty}\dfrac{8-3x}{9x^2+11}$ results in the indeterminate form $\dfrac{-\infty}{\infty}$. We should use l'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{8-3x}{9x^2+11} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[8-3x\right]}{\dfrac{d}{dx}[9x^2+11]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{-3}{18x} \\\\ &=\lim_{x\to\infty}\dfrac{-1}{6x} \\\\ &=0 \end{aligned}$ Note that we were only able to use l'Hôpital's rule because the limit $\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[8-3x\right]}{\dfrac{d}{dx}[9x^2+11]}$ can actually be determined. In conclusion, $\lim_{x\to\infty}\dfrac{8-3x}{9x^2+11}=0$.